= Solution
Use the <Euclidean path integral> with weight $e^{-S}$ and <Fourier transform> convention $\phi(x)=\int d^dp\,(2\pi)^{-d}e^{ip\cdot x}\phi(p)$. The momentum-space <Feynman rules> are: an internal <scalar propagator> $1/(p^2+m^2)$; a quartic vertex $-\lambda$ together with $(2\pi)^d\delta^{(d)}(\sum p_i)$ for incoming momenta; and an integral $\int d^dk/(2\pi)^d$ over each independent <loop momentum>. Multiply by the <Feynman-diagram symmetry factor>; external propagators are retained for a full <correlation function> and removed for an <amputated connected correlation function>.
To enumerate the requested <one-particle-irreducible Feynman diagrams>, let $V$ be the number of quartic vertices and $I$ the number of internal edges. Four external legs imply $4V=2I+4$, while one <loop order> implies $I-V+1=1$. Thus $V=I=2$. Both internal edges must connect the two vertices: an alternative with a tadpole and one connecting edge would disconnect on cutting that edge. The only graphs are therefore the three pairings of labelled external legs, each with <Feynman-diagram symmetry factor> $1/2$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-304-quartic-bubbles.png]
{title=The s, t and u one-loop one-particle-irreducible four-point graphs of quartic scalar theory}
The corresponding channel momenta are $q_s=p_1+p_2$, $q_t=p_1+p_3$, and $q_u=p_1+p_4$, with all external momenta incoming. This exhausts the connected one-loop four-point <one-particle-irreducible Feynman diagrams>.
For the <Euclidean massive loop integral>, first take $m^2>0$ and an integer $n>d/2$, where the integral converges. <Schwinger parameterization> gives
$$
\frac1{(k^2+m^2)^n}=\frac1{\Gamma(n)}\int_0^\infty ds\,s^{n-1}e^{-s(k^2+m^2)}.
$$
The supplied <Gaussian integral> then yields
$$
\int\frac{d^dk}{(2\pi)^d}e^{-sk^2}=(4\pi s)^{-d/2}.
$$
Integrating over $s$, and using the <Gamma function recurrence> to obtain $\Gamma(n)=(n-1)!$, proves
$$
\boxed{\int\frac{d^dk}{(2\pi)^d}\frac1{(k^2+m^2)^n}=\frac{(m^2)^{d/2-n}}{(4\pi)^{d/2}(n-1)!}\Gamma(n-d/2).}
$$
Outside its initial convergence range, the right-hand side defines the meromorphic continuation used in <dimensional regularization>; the divergent ordinary integral is not being assigned a convergent value.
For the <scalar bubble integral>, the <Feynman parameter> identity $1/(AB)=\int_0^1dx\,[xA+(1-x)B]^{-2}$ followed by a translation of <loop momentum> gives
$$
B(p)=\frac{\Gamma(2-d/2)}{(4\pi)^{d/2}}\int_0^1dx\,[m^2+x(1-x)p^2]^{d/2-2}.
$$
With $d=4-\epsilon$, the <Gamma function> satisfies $\Gamma(\epsilon/2)=2/\epsilon+O(1)$, and the parameter integral tends to one. Hence \b[the ultraviolet pole is independent of external momentum]:
$$
\boxed{B(p)=\frac1{8\pi^2\epsilon}+O(1).}
$$
One can also see why the same pole occurs without introducing a <Feynman parameter>: at large $k$, the difference between this integrand and $(k^2+m^2)^{-2}$ is ultraviolet integrable near four dimensions, so both have the same <dimensional regularization> pole. The positive mass avoids an infrared ambiguity in this argument.
In the <quantum effective action> convention, the tree-level four-point vertex is $+\lambda$, the negative of the <amputated connected correlation function> tree vertex. The three bubble corrections give
$$
\Gamma^{(4)}=\lambda-\frac{\lambda^2}{2}\{B(q_s)+B(q_t)+B(q_u)\}+\delta\lambda+O(\lambda^3).
$$
Thus \b[the pole counterterm] is $\delta\lambda=3\lambda^2/(16\pi^2\epsilon)$. The <modified minimal subtraction scheme> also subtracts the conventional finite $\log4\pi-\gamma_E$ combination, or equivalently absorbs it into the subtraction-scale convention; that does not change the one-loop <renormalization-group beta function>. Using that scale convention, write
$$
\lambda_B=\mu^\epsilon\left(\lambda+\frac{3\lambda^2}{16\pi^2\epsilon}+O(\lambda^3)\right).
$$
Here $\lambda$ is dimensionless, $[\lambda_B]=\epsilon$, and the conventional constant scale factor is implicit. The one-loop tadpole is momentum independent, so <wave-function renormalization> does not contribute at this order. Differentiating at fixed $\lambda_B$, with $a=3/(16\pi^2)$, gives $0=\epsilon(\lambda+a\lambda^2/\epsilon)+\beta(1+2a\lambda/\epsilon)+O(\lambda^3)$. Therefore
$$
\boxed{\beta(\lambda)=-\epsilon\lambda+\frac{3\lambda^2}{16\pi^2}+O(\lambda^3),\qquad\beta_{d=4}(\lambda)=\frac{3\lambda^2}{16\pi^2}+O(\lambda^3).}
$$
For the stable <quartic scalar field theory>, with $\lambda>0$, the <one-loop quartic scalar beta function> is positive: the <running coupling> increases toward the ultraviolet and decreases toward the infrared. The theory is not <asymptotically free>; extrapolating the one-loop flow gives a <Landau pole>.
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